Find the general solution of each of the following differential equations. If an initial condition is given, find the particular solution that satisfies this initial condition.

y' + 2ty = 1, y(0)=1

Question

Find the general solution of each of the following differential equations. If an initial condition is given, find the particular solution that satisfies this initial condition.

y' + 2ty = 1, y(0)=1

hy123 · Answer

@thomaster @RadEn

abdela25 · Answer

you want to find the value of y ??? 
if not could you please explain your question more because i cannot understand it very good

hy123 · Answer

Well it is a differential equation and i need to solve for y

hy123 · Answer

@ganeshie8

ganeshie8 · Answer

linear differential equation

ganeshie8 · Answer

start by finding the integrating factor

hy123 · Answer

Yes.

hy123 · Answer

Wouldn't that be. y' = 1 - 2ty?

hy123 · Answer

Making it a standard form and then diving both sides by y?

ganeshie8 · Answer

y' + p(x) y = q(x)

integrating factor =  $\large   e^{\int  p ~dx}$

hy123 · Answer

hm.. okay.

ganeshie8 · Answer

y' + 2ty = 1

integrating factor = $\large   e^{\int  2t ~dt} = e^{t^2}$

ganeshie8 · Answer

multiply that both sides

ganeshie8 · Answer

$\large   y' + 2ty = 1 $

multiply both sides wid the integrating factor $\large   e^{t^2}$

$\large   y'e^{t^2} + ye^{t^2}2t = e^{t^2} $

hy123 · Answer

Got that.

hy123 · Answer

Then do we take the integral of both sides?

ganeshie8 · Answer

yes, and observe that left side is simply a derivative of product :  $ye^{t^2}$

ganeshie8 · Answer

$\large   y' + 2ty = 1 $

multiply both sides wid the integrating factor $\large   e^{t^2}$

$\large   y'e^{t^2} + ye^{t^2}2t = e^{t^2} $

$\large  (ye^{t^2})' = e^{t^2} $

integrate both sides now..

hy123 · Answer

The integration of e^t^2 is a bit hard.

hy123 · Answer

Do you mind showing me how you got it.

ganeshie8 · Answer

nope, its simply a new funciton

hy123 · Answer

oh

hy123 · Answer

e^t^2 F(t) + C?

ganeshie8 · Answer

$\large  \int e^{x^2}dx =   \frac{\sqrt{\pi}}{2} erfi(x)$

hy123 · Answer

what is erfi?

ganeshie8 · Answer

$\large  \int e^{x^2}dx =   \frac{\sqrt{\pi}}{2} erfi(x) + C $

ganeshie8 · Answer

its a new function like logx and many other functions defined to be the area under the bell curve

hy123 · Answer

Ok.

ganeshie8 · Answer

$\large   y' + 2ty = 1 $

multiply both sides wid the integrating factor $\large   e^{t^2}$

$\large   y'e^{t^2} + ye^{t^2}2t = e^{t^2} $

$\large  (ye^{t^2})' = e^{t^2} $

integrate both sides now.. 

$\large   \int    (ye^{t^2})' = \int   e^{t^2} $

$\large   ye^{t^2} =  \frac{\sqrt{\pi}}{2} erfi(t) + C$

hy123 · Answer

Then you divide both side by e^t^2.

ganeshie8 · Answer

before that, find the value of C, 
apply initial conditions...

hy123 · Answer

Plugging in y(0) = 1 and solving for C?

ganeshie8 · Answer

yes

hy123 · Answer

I get C = -1/2

ganeshie8 · Answer

how ?

hy123 · Answer

Actually lemme re-do it.

ganeshie8 · Answer

okay

hy123 · Answer

Actually i got 1...

ganeshie8 · Answer

yes

ganeshie8 · Answer

plug that in and solve y

hy123 · Answer

C = 1? Correct?

ganeshie8 · Answer

correct

hy123 · Answer

Do you mind showing me what you got?

hy123 · Answer

Nvm i got it. Now i got to hed to class. Thanks a lot Ganeshie.

ganeshie8 · Answer

$\large   y' + 2ty = 1 $

multiply both sides wid the integrating factor $\large   e^{t^2}$

$\large   y'e^{t^2} + ye^{t^2}2t = e^{t^2} $

$\large  (ye^{t^2})' = e^{t^2} $

integrate both sides now.. 

$\large   \int    (ye^{t^2})' = \int   e^{t^2} $

$\large   ye^{t^2} =  \frac{\sqrt{\pi}}{2} erfi(t) + C$  *****

y(0) = 1

$\large   1e^{0} =  \frac{\sqrt{\pi}}{2} erfi(0) + C$

$\large   1 =  0 + C$

$\large   1 =  C$



$\large   ye^{t^2} =  \frac{\sqrt{\pi}}{2} erfi(t) + 1$ 

$\large   y =  e^{-t^2} ( \frac{\sqrt{\pi}}{2} erfi(t) + 1)$

ganeshie8 · Answer

u wlc :)